# LOOP problem with DIAGRAM!

1. Nov 29, 2007

### physicsbhelp

[SOLVED] LOOP problem with DIAGRAM!

A mass (m) slides without friction down a ramp and around a loop. What is the minimum height that the block can be released from to make it around the loop?
i have ATTACHED the picture of the diagram!! in the picture the R is the radius, the m is the mass, and the h is the height.
i would appreciate any help on this problem.

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Last edited: Nov 29, 2007
2. Nov 29, 2007

### alvaros

1 - You have to calculate the velocity the mass needs "to make it around the loop". When the mass is at the top of the loop its centrifugal acceleration must be greater than gravity. That depends just on R
2 - Using conservation of energy you have to calculate the velocity the mass will have when it is at the top of the loop. That depends just on ( minimum_height - R )

3. Nov 29, 2007

### physicsbhelp

thanks, but im still very confused because our physics teacher didnt give us any vlues for these variables. also i dont understand the centrifugal acceleration?

4. Nov 29, 2007

### Staff: Mentor

What is the graviational potential energy at h?

What is the difference in gravitational potential energy between the bottom of the loop and the top?

Thinking about conservation of energy, are there any other forms of energy involved in this problem?

5. Nov 29, 2007

### physicsbhelp

well i interpret your answer as asking me questions, but just to let you know the problem only gave the things that i said so it doesnt tell me the things you asked me. but isnt the Ug=mgh and so that is the "graviational potential energy at h" right?

and for 2) "What is the difference in gravitational potential energy between the bottom of the loop and the top?"
well the top would probably have mgh + mv(squared)/ 2 but the bottom would not have any potential energy becuase there is no height so its energy would be just mv(squared)/ 2, right?

3) "Thinking about conservation of energy, are there any other forms of energy involved in this problem?"
the other form of energy involved could be Work? maybe i dont know??

6. Nov 29, 2007

### physicsbhelp

PLEASE HELP, any help will be greatly appreciated. this is one of the most challenging problems i have ever done at my level of physics.

7. Nov 29, 2007

### Staff: Mentor

Correct!

Now what is the difference in elevation between the bottom and top of the loop?

8. Nov 29, 2007

### physicsbhelp

i think the difference would be diameter

9. Nov 29, 2007

### physicsbhelp

is that right?

10. Nov 29, 2007

### physicsbhelp

but after i do that i still don't see how to get height?>?

11. Nov 29, 2007

### Staff: Mentor

Correct.

So falling height h changes the GPE by mgh, which becomes kinetic energy if no dissipative forces are involved (mass slides without friction). The process is reversed as the mass climbs 2R in the loop, and it must have just enough tangential velocity to make it pass through the top of the loop.

Let's say the mass started at top of the loop and it slid without friction or air resistance. What would happen?

12. Nov 29, 2007

### physicsbhelp

first off what is tangential velocity?

secondly, for your questions i think that the mass block would just fall off. right?

13. Nov 29, 2007

### Staff: Mentor

OK, I may have misunderstood your problem, looking back at alvaros's post. I thought the mass was perhaps tied to the track, and it only need to make it to the top of the loop, in which case h = 2R.

Now based on your last post (and reflecting on "secondly, for your questions i think that the mass block would just fall off. right?") and reflecting on the post of alvaros, the mass is slimply sliding on top of the guide, and is not physically tied to the wire, so as alvaros correctly pointed out the centrifugal force of the mass mv2/R must = mg for the mass to be weightless at the top of the loop. v is the tangential velocity, i.e. the linear velocity tangent to the loop.

So mg = mv2/R for a weightless condition. The kinetic energy associated with this tangential velocity must come from the change in gravitational potential energy with the mass above and elevation of 2R.

Think of h = h1 + h2. One needs to find h1 above 2R, since h2 = 2R.

14. Nov 29, 2007

### physicsbhelp

i think you are still misunderstanidng my diagram. the line in the circle is the raidus but there is no wire there or anything like that. that line is there to clairfy that that is the radius. but i dont get how h2 = 2r and are you multiplying h by 2 to get h2?

15. Nov 29, 2007

### physicsbhelp

okay so is h = (mgR) / 2 + 2R ?????????
and the question that the problem asks:What is the minimum height that the block can be released from to make it around the loop? and just to confirm with you that does mean that in my diagram we are trying to find the h from the mass at the top of the slope and the ground...right?

16. Nov 29, 2007

### Staff: Mentor

Sorry for the confusion with the symbols/variables.

should read as h = h1 + h2, where h2 = 2R.

Assuming conservation of energy, the mass falls a height h, to the base of the loop, so the total energy mgh is converted to kinetic energy at that point. Then the mass climbs 2R (h2 in my previous example), so the kinetic energy decreases by mg(2R).

At the top of the loop, the KE = 1/2 mv2, where v is the tangential velocity, which must satisfy the condition for m to be weightless so that it will not fall off the track. That condition is mg = mv2/R.

So mgh = 1/2 mv2 + mg (2R) gives the minimum h.

17. Nov 29, 2007

### physicsbhelp

i really dont know how to solve that equation that you gave. but one more question when the mass block reaches the top of the loop i think there is both PE and KE there. right?
and how do you go about solving the equation? i have been trying would you first divide both sides by mg?

18. Nov 29, 2007

### Staff: Mentor

h = R/2 + 2R (making sure units are consistent), and yes h is the height from the common reference elevation which is the base of the loop.

19. Nov 29, 2007

### physicsbhelp

so i think we get h = 1/2 vsquared + 2R

20. Nov 29, 2007

### Staff: Mentor

Well, I think you did to get where you got, but something was amiss.

Taking - mgh = 1/2 mv2 + mg (2R)

and using mg = mv2/R, one gets v2 = gR

mgh = 1/2 m (gR) + mg (2R), and then one can divide by mg.

h = 1/2 R + 2R